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Class 10 Mathematics Unit 1 & Unit 2 Model Paper 2027 | Complex Numbers, Quadratic Equations and Inequalities | Board Standard
Class 10 Math Model Paper 2027
Class 10 Mathematics Model Paper 2027 covering Unit 1: Complex Numbers and Unit 2: Quadratic Equations and Inequalities is designed according to a board-standard paper pattern. This Class 10 Mathematics Model Paper 2027 includes objective and subjective-type questions, short questions, and detailed questions based on important concepts from Complex Numbers and Quadratic Equations and Inequalities. It is useful for Class 10 Mathematics exam preparation, revision, practice tests, and students preparing for the 2027 Board Examination.
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PAPER NO.
01
MATHEMATICS
CHAPTER
WISE
Roll No. (in Figures):
(in Words):
Unit 1: Complex Numbers
Unit 2: Quadratic Equations and Inequalities
Unit 2: Quadratic Equations and Inequalities
OBJECTIVE TYPE
Time : 20 Minutes
Marks : 15
Marks : 15
1ABCD
2ABCD
3ABCD
4ABCD
5ABCD
6ABCD
7ABCD
8ABCD
9ABCD
10ABCD
11ABCD
12ABCD
13ABCD
14ABCD
15ABCD
Note: Four possible answers A, B, C and D to each question are given.
The choice which you think is correct, fill that circle in front of that question with
Marker or Pen ink. Cutting or filling two or more circles will result in zero mark in that question.
Q1. 15
1. Real part of (2 − 3i)(2 + 3i) is:
Answer: (D) 13
- The product has the form (a − bi)(a + bi) = a² − b²i².
- Here a = 2, b = 3, so (2)² − (3)²i² = 4 − 9i².
- Since i² = −1: 4 − 9(−1) = 4 + 9 = 13.
- So the number is 13 + 0i, and its real part is 13.
2. What is additive inverse of 5 − 2i?
Answer: (D) −5 + 2i
- The additive inverse of z is the number that gives 0 when added to z, i.e. −z.
- −(5 − 2i) = −5 + 2i.
- Check: (5 − 2i) + (−5 + 2i) = 0 ✔
3. If z = 4 − 3i, then z z̄ =
Answer: (D) 25
- z = 4 − 3i, so the conjugate z̄ = 4 + 3i.
- z z̄ = (4 − 3i)(4 + 3i) = 4² − (3i)² = 16 − 9i².
- Since i² = −1: 16 + 9 = 25.
- Shortcut: z z̄ = a² + b² = 4² + 3² = 25.
4. Conjugate of 9 − 4i is:
Answer: (B) 9 + 4i
- The conjugate of a + bi is a − bi, i.e. only the sign of the imaginary part changes.
- For 9 − 4i, keep the real part 9 and change −4i to +4i.
- Conjugate = 9 + 4i.
5. The imaginary unit is represented by:
Answer: (B) i
- The imaginary unit is defined as i = √(−1), so i² = −1.
- α and β are usually used for roots of an equation, and ω for a cube root of unity.
- Hence the imaginary unit is i.
6. The idea of complex numbers first emerged in:
Answer: (B) 16th century
- This is a history-based question.
- The Italian mathematician Gerolamo Cardano (1545) worked with square roots of negative numbers while solving cubic equations.
- Rafael Bombelli later set out rules for them. Both lived in the 16th century.
7. (4 − 3i) − (2 − 5i) = ______
Answer: (A) 2 + 2i
- Remove the brackets, changing the signs of the second bracket: 4 − 3i − 2 + 5i.
- Real parts: 4 − 2 = 2.
- Imaginary parts: −3i + 5i = 2i.
- Result = 2 + 2i.
8. The number (2 − 7i) is equivalent to:
Answer: (A) (2, −7)
- A complex number a + bi can be written as the ordered pair (a, b).
- For 2 − 7i: a = 2 and b = −7.
- So 2 − 7i ≡ (2, −7).
9. The solution set of 3x² − 9 = 0 is:
Answer: (C) {±√3}
- 3x² − 9 = 0 ⟹ 3x² = 9.
- Divide by 3: x² = 3.
- Take the square root of both sides: x = ±√3.
- Solution set = {±√3}.
10. Product of the roots of 3x² + 5x − 12 = 0 is:
Answer: (A) −4
- Compare with ax² + bx + c = 0: a = 3, b = 5, c = −12.
- For a quadratic, product of roots = c/a.
- Product = −12 / 3 = −4.
11. 3 and 2 are the roots of:
Answer: (C) x² − 5x + 6 = 0
- Equation with roots α, β: x² − (α + β)x + αβ = 0.
- Sum = 3 + 2 = 5 and product = 3 × 2 = 6.
- Equation: x² − 5x + 6 = 0.
- Check: (x − 3)(x − 2) = x² − 5x + 6 ✔
12. If b² − 4ac > 0 and is a perfect square, then the roots of ax² + bx + c = 0 are:
Answer: (B) unequal
- The roots are x = (−b ± √(b² − 4ac)) / 2a.
- Since b² − 4ac > 0, the square root is real and non-zero, so the two roots are unequal and real.
- Since b² − 4ac is also a perfect square, √(b² − 4ac) is a rational number, so the roots are rational as well (rational and unequal).
- Option (D) irrational is wrong because irrational roots occur only when the discriminant is not a perfect square. Among the options, the correct one is unequal.
13. An equation of the form ax² + bx + c = 0 becomes pure quadratic equation when:
Answer: (D) b = 0
- A pure quadratic equation has only the x² term and a constant, with no x term: ax² + c = 0.
- The x term is bx, so it disappears when b = 0.
- If a = 0 the equation would not be quadratic at all, and if c = 0 it becomes ax² + bx = 0 (an adfected quadratic).
14. The point of intersection of y = 2x + 4 with x-axis graphically is:
Answer: (B) (−2, 0)
- Every point on the x-axis has y = 0.
- Put y = 0: 0 = 2x + 4 ⟹ 2x = −4.
- x = −2.
- The line meets the x-axis at (−2, 0).
15. The expression b² − 4ac in a quadratic equation is known as:
Answer: (D) discriminant
- In x = (−b ± √(b² − 4ac)) / 2a, the expression under the root decides the nature of the roots.
- It “discriminates” between real/imaginary and equal/unequal roots, so it is called the discriminant.
Here is subjective Part of Class 10 th Model Paper from Unit No.No.1 and it No. 2
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Roll No. (in Figures):
(in Words):
Unit 1: Complex Numbers
Unit 2: Quadratic Equations and Inequalities
Unit 2: Quadratic Equations and Inequalities
Time : 2:10 Hours
Marks: 60
Marks: 60
SUBJECTIVE TYPE (PART - I)
Q2. Write short answers to any SIX (6) questions:(6×2=12)
(i) Is "0" a complex number? Explain.
Ans: Yes. Write 0 = 0 + 0i. Since a + bi with a = 0, b = 0 is of the complex form, 0 is a complex number (real part 0, imaginary part 0).
(ii) State the condition for two complex numbers to be equal.
Ans: a + bi = c + di if and only if a = c (real parts equal) and b = d (imaginary parts equal).
(iii) Simplify: (3 − 4i)(5 − 6i)
Sol: = 15 − 18i − 20i + 24i²
= 15 − 38i − 24 (∵ i² = −1)
= −9 − 38i
= 15 − 38i − 24 (∵ i² = −1)
= −9 − 38i
(iv) If z₁ = 5 + 4i, and z₂ = 3 + 2i, then find z₁z₂.
Sol: z₁z₂ = (5 + 4i)(3 + 2i) = 15 + 10i + 12i + 8i²
= 15 + 22i − 8 = 7 + 22i
= 15 + 22i − 8 = 7 + 22i
(v) Find real and imaginary parts of z = (2 + 7i)⁻¹.
Sol: z = 1/(2 + 7i) × (2 − 7i)/(2 − 7i) = (2 − 7i)/(4 + 49) = (2 − 7i)/53
= 2/53 − (7/53)i
Real part = 2/53, Imaginary part = −7/53
= 2/53 − (7/53)i
Real part = 2/53, Imaginary part = −7/53
(vi) Why does the equation x² + 1 = 0 has no "real" solution?
Ans: x² + 1 = 0 gives x² = −1. The square of every real number is non-negative, so no real number satisfies it. Its roots are x = ±i (imaginary).
(vii) Find additive and multiplicative inverse of z = 8 + 9i.
Sol: Additive inverse = −z = −8 − 9i
Multiplicative inverse = 1/z = (8 − 9i)/((8)² + (9)²) = (8 − 9i)/145
= 8/145 − (9/145)i
Multiplicative inverse = 1/z = (8 − 9i)/((8)² + (9)²) = (8 − 9i)/145
= 8/145 − (9/145)i
(viii) Simplify: (17 − 7i) + (19 − 7i)
Sol: = (17 + 19) + (−7 − 7)i = 36 − 14i
(ix) If z₁ + z₂ = 19 − 11i and z₁ = 3 − 7i then find z₂.
Sol: z₂ = (z₁ + z₂) − z₁ = (19 − 11i) − (3 − 7i)
= (19 − 3) + (−11 + 7)i = 16 − 4i
= (19 − 3) + (−11 + 7)i = 16 − 4i
Q3. Write short answers to any SIX (6) questions:(6×2=12)
(i) Express (2 + √3i)(−7 + 2i) in the form of a + bi.
Sol: = −14 + 4i − 7√3 i + 2√3 i²
= −14 − 2√3 + (4 − 7√3)i
= (−14 − 2√3) + (4 − 7√3)i
= −14 − 2√3 + (4 − 7√3)i
= (−14 − 2√3) + (4 − 7√3)i
(ii) If z₁ = 2 + 3i and z₂ = 4 + i, show that the commutative property of addition holds.
Sol: z₁ + z₂ = (2 + 3i) + (4 + i) = 6 + 4i
z₂ + z₁ = (4 + i) + (2 + 3i) = 6 + 4i
Hence z₁ + z₂ = z₂ + z₁. Proved.
z₂ + z₁ = (4 + i) + (2 + 3i) = 6 + 4i
Hence z₁ + z₂ = z₂ + z₁. Proved.
(iii) Write the Associative property for multiplication of complex numbers.
Ans: For all z₁, z₂, z₃ ∈ ℂ: z₁(z₂z₃) = (z₁z₂)z₃
(iv) What is an Argand Plane?
Ans: The plane in which the horizontal axis (x-axis) represents the real part and the vertical axis (y-axis) represents the imaginary part. Each complex number a + bi is shown as the point (a, b). It is also called the complex plane.
(v) Examine the nature of the roots of the equation: 15x² + 11x + 2 = 0.
Sol: a = 15, b = 11, c = 2
Disc = b² − 4ac = 121 − 4(15)(2) = 121 − 120 = 1
Disc > 0 and a perfect square, so roots are real, rational and unequal.
Disc = b² − 4ac = 121 − 4(15)(2) = 121 − 120 = 1
Disc > 0 and a perfect square, so roots are real, rational and unequal.
(vi) Examine the nature of the roots of the equation: x² − x − 1 = 0.
Sol: a = 1, b = −1, c = −1
Disc = (−1)² − 4(1)(−1) = 1 + 4 = 5
Disc > 0 but not a perfect square, so roots are real, irrational and unequal.
Disc = (−1)² − 4(1)(−1) = 1 + 4 = 5
Disc > 0 but not a perfect square, so roots are real, irrational and unequal.
(vii) If a ball is thrown upward with a velocity v, the maximum height it reaches can be determined by using a formula h = v²/2g. Rearrange the formula to make v the subject.
Sol: h = v²/(2g)
Multiply both sides by 2g: v² = 2gh
Take square root: v = √(2gh)
Multiply both sides by 2g: v² = 2gh
Take square root: v = √(2gh)
(viii) What is the standard form of a quadratic equation?
Ans: ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0.
(ix) Solve the equation x² + 2x − 35 = 0 by factorization method.
Sol: Find two numbers with product −35 and sum 2: 7 and −5.
x² + 7x − 5x − 35 = 0
x(x + 7) − 5(x + 7) = 0 → (x + 7)(x − 5) = 0
x = −7 or x = 5. S.S = {−7, 5}
x² + 7x − 5x − 35 = 0
x(x + 7) − 5(x + 7) = 0 → (x + 7)(x − 5) = 0
x = −7 or x = 5. S.S = {−7, 5}
Q4. Write short answers to any SIX (6) questions:(6×2=12)
(i) Solve the equation by completing square method: x² − 2x − 889 = 0.
Sol: x² − 2x = 889
Add (½ × 2)² = 1 on both sides: x² − 2x + 1 = 890
(x − 1)² = 890 → x − 1 = ±√890
x = 1 ± √890. S.S = {1 + √890, 1 − √890}
Add (½ × 2)² = 1 on both sides: x² − 2x + 1 = 890
(x − 1)² = 890 → x − 1 = ±√890
x = 1 ± √890. S.S = {1 + √890, 1 − √890}
(ii) Write the formula for the sum of the roots of a quadratic equation.
Ans: For ax² + bx + c = 0: α + β = −b/a
(iii) What happens to the product of the roots if constant term c is zero?
Ans: Product of roots = c/a = 0/a = 0. So the product is zero, meaning one root is zero.
(iv) What is the discriminant of the equation ax² + bx + c = 0?
Ans: Discriminant = b² − 4ac
(v) Describe the nature of roots if b² − 4ac > 0 and is a perfect square.
Ans: The roots are real, rational and unequal (given a, b, c are rational).
(vi) Examine the nature of the roots 25x² − 30x + 9 = 0.
Sol: a = 25, b = −30, c = 9
Disc = 900 − 4(25)(9) = 900 − 900 = 0
Disc = 0, so roots are real, rational and equal (x = 3/5, 3/5).
Disc = 900 − 4(25)(9) = 900 − 900 = 0
Disc = 0, so roots are real, rational and equal (x = 3/5, 3/5).
(vii) Examine the nature of the roots of x² − 7x + 12 = 0.
Sol: a = 1, b = −7, c = 12
Disc = 49 − 48 = 1 (positive perfect square)
Roots are real, rational and unequal (x = 3, 4).
Disc = 49 − 48 = 1 (positive perfect square)
Roots are real, rational and unequal (x = 3, 4).
(viii) If y² = 4ax, make 'a' as the subject of the equation.
Sol: Divide both sides by 4x: a = y²/(4x)
(ix) If I = PRT then make 'R' the subject of this formula.
Sol: Divide both sides by PT: R = I/(PT)
(PART - II)
Note: Attempt any two (02) questions.(2×8=16)
Q5. (a) If (1 + i)² / (2 − i) = x + iy then find the values of x and y.
Sol:
Step 1: (1 + i)² = 1 + 2i + i² = 1 + 2i − 1 = 2i
Step 2: LHS = 2i/(2 − i)
Step 3: Multiply by conjugate (2 + i)/(2 + i):
= 2i(2 + i)/((2)² + (1)²) = (4i + 2i²)/5 = (−2 + 4i)/5
Step 4: x + iy = −2/5 + (4/5)i
Comparing: x = −2/5, y = 4/5
Step 1: (1 + i)² = 1 + 2i + i² = 1 + 2i − 1 = 2i
Step 2: LHS = 2i/(2 − i)
Step 3: Multiply by conjugate (2 + i)/(2 + i):
= 2i(2 + i)/((2)² + (1)²) = (4i + 2i²)/5 = (−2 + 4i)/5
Step 4: x + iy = −2/5 + (4/5)i
Comparing: x = −2/5, y = 4/5
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(b) If (2x + iy)(1 − i) = 4 + 2i, then find the values of x and y.
Sol:
Step 1: Expand: 2x − 2xi + iy − i²y = 2x + y + (y − 2x)i
Step 2: (2x + y) + (y − 2x)i = 4 + 2i
Step 3: Compare: 2x + y = 4 …(1) and −2x + y = 2 …(2)
Step 4: Add (1) and (2): 2y = 6 → y = 3
Step 5: From (1): 2x = 4 − 3 = 1 → x = 1/2
x = 1/2, y = 3
Step 1: Expand: 2x − 2xi + iy − i²y = 2x + y + (y − 2x)i
Step 2: (2x + y) + (y − 2x)i = 4 + 2i
Step 3: Compare: 2x + y = 4 …(1) and −2x + y = 2 …(2)
Step 4: Add (1) and (2): 2y = 6 → y = 3
Step 5: From (1): 2x = 4 − 3 = 1 → x = 1/2
x = 1/2, y = 3
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Q6. (a) If z = 4 − 3i, then verify that |z| = |−z| = |z̄| = |−z̄|.
Sol: z = 4 − 3i, −z = −4 + 3i, z̄ = 4 + 3i, −z̄ = −4 − 3i
|z| = √(16 + 9) = √25 = 5
|−z| = √((−4)² + 3²) = √25 = 5
|z̄| = √(4² + 3²) = √25 = 5
|−z̄| = √((−4)² + (−3)²) = √25 = 5
Hence |z| = |−z| = |z̄| = |−z̄| = 5. Verified.
|z| = √(16 + 9) = √25 = 5
|−z| = √((−4)² + 3²) = √25 = 5
|z̄| = √(4² + 3²) = √25 = 5
|−z̄| = √((−4)² + (−3)²) = √25 = 5
Hence |z| = |−z| = |z̄| = |−z̄| = 5. Verified.
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(b) Find real and imaginary parts of (4 + 3i)⁻¹.
Sol:
(4 + 3i)⁻¹ = 1/(4 + 3i)
Multiply by conjugate: (4 − 3i)/((4 + 3i)(4 − 3i)) = (4 − 3i)/(16 + 9) = (4 − 3i)/25
= 4/25 − (3/25)i
Real part = 4/25, Imaginary part = −3/25
(4 + 3i)⁻¹ = 1/(4 + 3i)
Multiply by conjugate: (4 − 3i)/((4 + 3i)(4 − 3i)) = (4 − 3i)/(16 + 9) = (4 − 3i)/25
= 4/25 − (3/25)i
Real part = 4/25, Imaginary part = −3/25
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Q7. (a) Find real and imaginary parts of z = (4 + 3i)⁻².
Sol:
Step 1: z = 1/(4 + 3i)²
Step 2: (4 + 3i)² = 16 + 24i + 9i² = 7 + 24i
Step 3: z = 1/(7 + 24i) = (7 − 24i)/(49 + 576) = (7 − 24i)/625
Step 4: z = 7/625 − (24/625)i
Real part = 7/625, Imaginary part = −24/625
Step 1: z = 1/(4 + 3i)²
Step 2: (4 + 3i)² = 16 + 24i + 9i² = 7 + 24i
Step 3: z = 1/(7 + 24i) = (7 − 24i)/(49 + 576) = (7 − 24i)/625
Step 4: z = 7/625 − (24/625)i
Real part = 7/625, Imaginary part = −24/625
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(b) Find the sum and the product of the roots of the equation 3x² + 5x − 12 = 0 without solving.
Sol: a = 3, b = 5, c = −12
Sum of roots = −b/a = −5/3
Product of roots = c/a = −12/3 = −4
Sum of roots = −b/a = −5/3
Product of roots = c/a = −12/3 = −4
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(PART - III)
Note: Attempt any one (01) question.(1×8=8)
Q8. (a) Find the condition that the roots of ax² + bx + c = 0 may be equal in magnitude but opposite in sign, where a ≠ 0.
Sol:
Let the roots be α and −α.
Sum of roots = α + (−α) = 0
But sum of roots = −b/a
So −b/a = 0 → b = 0
Condition: b = 0 (and a, c of opposite signs for real roots).
Let the roots be α and −α.
Sum of roots = α + (−α) = 0
But sum of roots = −b/a
So −b/a = 0 → b = 0
Condition: b = 0 (and a, c of opposite signs for real roots).
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(b) If the quadratic equation 16x² + 7px + 49 = 0 has equal roots, then find the values of p.
Sol: a = 16, b = 7p, c = 49
Equal roots → b² − 4ac = 0
(7p)² − 4(16)(49) = 0
49p² − 3136 = 0 → p² = 3136/49 = 64
p = ±8
Equal roots → b² − 4ac = 0
(7p)² − 4(16)(49) = 0
49p² − 3136 = 0 → p² = 3136/49 = 64
p = ±8
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Q9. (a) Solve x² − 5x + 2 = 0 by using the quadratic formula.
Sol: a = 1, b = −5, c = 2
x = (−b ± √(b² − 4ac))/2a = (5 ± √(25 − 8))/2
x = (5 ± √17)/2
S.S = {(5 + √17)/2, (5 − √17)/2}
x = (−b ± √(b² − 4ac))/2a = (5 ± √(25 − 8))/2
x = (5 ± √17)/2
S.S = {(5 + √17)/2, (5 − √17)/2}
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(b) Find the point of intersecting of the linear equation y = 2 + 3x with coordinate axes graphically.
Sol:
Step 1: Table of values for y = 3x + 2:
x = −1 → y = −1 | x = 0 → y = 2 | x = 1 → y = 5
Step 2: Plot (−1, −1), (0, 2), (1, 5) and join with a straight line.
Step 3: y-axis intercept (put x = 0): y = 2 → point (0, 2)
Step 4: x-axis intercept (put y = 0): 0 = 2 + 3x → x = −2/3 → point (−2/3, 0)
The line cuts the x-axis at (−2/3, 0) and the y-axis at (0, 2).
Step 1: Table of values for y = 3x + 2:
x = −1 → y = −1 | x = 0 → y = 2 | x = 1 → y = 5
Step 2: Plot (−1, −1), (0, 2), (1, 5) and join with a straight line.
Step 3: y-axis intercept (put x = 0): y = 2 → point (0, 2)
Step 4: x-axis intercept (put y = 0): 0 = 2 + 3x → x = −2/3 → point (−2/3, 0)
The line cuts the x-axis at (−2/3, 0) and the y-axis at (0, 2).
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